Find a polynomial over $\mathbb{Q}$ with a given Galois group. 
How can I find a polynomial over $\mathbb{Q}$ given its Galois group over $\mathbb{Q}$. 

Do I have to guess, i.e. pick random polynomials and calculate their Galois groups, or is there a clever method?
I'm asking this, because I have an assignment that asks to find a polynomial over $\mathbb{Q}$, such that its Galois group is $\mathbb{Z}_3\times\mathbb{Z}_3$ over $\mathbb{Q}$. 
 A: For a given finite abelian group $ G $, there is a relatively straightforward algorithm which produces a Galois extension $ L/\mathbf Q $ with Galois group $ G $. Then, the minimal polynomial of any primitive element of this extension has Galois group $ G $ over $ \mathbf Q $.
For this algorithm, use the structure theorem for finite abelian groups to write
$$ G \cong C_{n_1} \times C_{n_2} \times \ldots \times C_{n_k} $$ 
for some $ n_i $. Now, find distinct prime numbers $ p_1, p_2, \ldots, p_k $ such that $ p_i \equiv 1 \pmod{n_i} $ for each $ i $. (Such primes always exist by Dirichlet's theorem on arithmetic progressions.) Let $ d = p_1 p_2 \ldots p_k $, and let $ \zeta $ be a primitive $ d $th root of unity. Then, we have that 
$$ \textrm{Gal}(\mathbf Q(\zeta_d)/\mathbf Q) \cong (\mathbf Z/d\mathbf Z)^{\times} \cong \prod_{i=1}^k (\mathbf Z/p_i \mathbf Z)^{\times} \cong \prod_{i=1}^k C_{p_i - 1} $$
Now, quotienting this group by the subgroup
$$ H = \prod_{i=1}^k C_{(p_i - 1)/n_i} $$
gives the group $ G $, and quotienting by $ H $ is equivalent to finding the Galois group of the fixed field of $ H $ over $ \mathbf Q $. Therefore, the fixed field of $ H $ in $ \mathbf Q(\zeta_d) $ is a Galois extension of $ \mathbf Q $ with Galois group $ G $.
Let's see how this works in practice. We want to find a Galois extension of $ \mathbf Q $ with Galois group $ C_3 \times C_3 $, so we want two distinct primes which are both $ 1 $ modulo $ 3 $. We may choose $ p_1 = 7 $ and $ p_2 = 13 $, so that $ d = 91 $. The fixed field of $ H $ is then the compositum of the cubic subfields of $ \mathbf Q(\zeta_7) $ and $ \mathbf Q(\zeta_{13}) $. With some help from Wolfram and the normal basis theorem, we may find that these cubic subfields are the splitting fields of $ X^3 + X^2 - 2X - 1 $ and $ X^3 + X^2 - 4X + 1 $, respectively. Therefore, the product
$$ (X^3 + X^2 - 2X - 1)(X^3 + X^2 - 4X + 1) $$
is a polynomial with Galois group $ C_3 \times C_3 $ over $ \mathbf Q $.
